Φ21
Fall 2006
Section 210 (8am)
Quiz 3 Solutions
Cyclotron Motion
Consider an electron (me = 9.11 × 10−31 kg, qe = −1.6 × 10−19 C) moving perpendicular to a magnetic eld
with a magnitude B = 3.402 mT. The magnetic eld will cause the electron to travel in a circular path
without changing its speed.
If the electron is moving at v1 = 1.5 × 105 m/s, what is the radius r1 of the cyclotron motion? In
this case, how many times per second f1 does the electron orbit?
Part 1.
Solution:
251 µm.
~ , F = |q| vB =
Because ~v ⊥ B
The number of orbits per second is f =
mv 2
r
1
∆t
, and solving for r gives r =
=
v
C
=
v
2πr
=
|q|B
2πm
mv
|q|B
=
(9.11×10−31 kg)(1.5×105 m/s)
(1.6×10−19 C)(3.402×10−3 T)
=
= 9.51 × 107 Hz = 95.1 MHz.
If the electron is moving at v2 = 2 × 105 m/s, what is the radius r2 of the cyclotron motion? In this
case, how many times per second f2 does the electron orbit?
Part 2.
Solution: Using the same formulas with a dierent v gives r2 = 335 nm and f2 = 95.1 GHz. The frequency
is unchanged!
~ eld is up, in what direction
If the negatively-charged electron is travelling to the right, and the B
will the electron rst be pulled? (Choose from left, right, up, down, into the page, out of the page.)
Part 3.
~ up, the ~v × B
~ term is out of the page. Because this is an electron,
Solution: With motion to the right and B
the direction of the force is opposite, or INTO THE PAGE.
Equations
~
F~ = q~v × B
F = ma
a=
1
v2
r
C = 2πr
v = ∆s/∆t